Math 1021, Linear Algebra 1. Section: A at 10am, B at 2:30pm

Math 1021, Linear Algebra 1. Section: A at 10am, B at 2:30pm All course information is available on Moodle. Text: Nicholson, Linear algebra with applications, 7th edition. We shall cover Chapters 1,2,3,4,5: 1. Systems of Linear Equations: 1.1 1.3 2. Matrix Algebra: 2.1 2.7 3. Determinants and Diagonalization: 3.1 3.4 4. Vector Geometry: 4.1 4.4 5. The Vector Space R n : 5.1 5.6 Evaluation: Weekly online homeworks: 10% (registration on Lyryx is free, this year only). Test 1: Tuesday, October 10, in class, 25% (Note the date change!) Test 2: Thursday, November 23, in class, 25% Final Exam: 40%

Suggested exercises from Nicholson s Linear algebra with applications, 7th ed. are available on Moodle. Important: The online Lyryx homework is a good technical drill. As such it is not a sufficient preparation for the tests and the exam. Working out the suggested exercises will provide a good preparation for the tests and the exam. The following strategy: 1. Read an exercise. 2. Read its solution without stopping to think about the exercise. 3. Proceed to the next exercise immediately.

1.1 Systems of Linear Equations, quick recap A linear equation in the n variables x 1,..., x n is a 1 x 1 +... a n x n = b. a 1,..., a n are coefficients of x 1,..., x n. b is the constant term in the equation. A system of linear equations in variables x 1,..., x n is a finite collection of linear equations in these variables. A system of three equations in four variables looks like this: a 11 x 1 + a 12 x 2 + a 13 x 3 + a 14 x 4 = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 + a 24 x 4 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 + a 34 x 4 = b 3

A solution of a system of linear equations in variables x 1,..., x n is a sequence of n real numbers s 1,..., s n which is a solution to every equation in the system. The basic problem of linear algebra: Describe and study an algorithm for solving systems on linear equations. Our goal #1: Describe an algorithm that checks whether a given system has a solution. If it does, the algorithm describes the solution in parametric form.

Given a system Augmented matrix of a system a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 of three equations in three variables. The array of numbers a 11 a 12 a 13 b 1 a 21 a 22 a 23 b 2 a 31 a 32 a 33 b 3 is the augmented matrix of the system.

1.2 Gaussian elimination Example Which of the following three systems is easiest to solve? 2x + y + 3z = 1 2y z + x = 0 9z + x 4y = 2 x + y + 3z = 1 y z = 0 z = 2 x = 1 y = 0 z = 2 (1) (2) (3)

A useful notational convention In the following * stands for any real number, possibly zero. So [ ] 1 0 may stand for [ 1 2 3 0 ] [ or 1 30 30 0 ] [ 0 2, but not 3 0 ].

The row-echelon form (REF) and the reduced row-echelon form (RREF) Definition A matrix is in the row-echelon form (REF) if it satisfies the following: 1. All zero rows (if any) are at the bottom. 2. The first nonzero entry in each nonzero row is a 1. It is called the leading 1 for that row. 3. Each leading 1 is to the right of all leading 1s in the rows above it. An REF matrix is said to be in the reduced row-echelon form (RREF) if in addition to the above conditions 4. Each leading 1 is the only nonzero entry in its column.

Solving systems whose augmented matrix is in the RREF This is a very short slide! If the matrix is (writing a, b, c,... in place of *s) 1 0 a 0 = b 0 1 c 0 = d 0 0 0 1 = e then the system is x 1 + ax 3 = b x 2 + cx 3 = d x 4 = e

Solving systems whose augmented matrix is in the RREF This is a very short slide! If the matrix is (writing a, b, c,... in place of *s) 1 0 a 0 = b 0 1 c 0 = d 0 0 0 1 = e then the system is or x 1 + ax 3 = b x 2 + cx 3 = d x 4 = e x 1 = b ax 3 x 2 = d cx 3 x 4 = e

Solving systems whose augmented matrix is in the RREF This is a very short slide! If the matrix is (writing a, b, c,... in place of *s) 1 0 a 0 = b 0 1 c 0 = d 0 0 0 1 = e then the system is or and the general solution is x 1 + ax 3 = b x 2 + cx 3 = d x 4 = e x 1 = b ax 3 x 2 = d cx 3 x 4 = e x 1 = b ap, x 2 = d cp, x 3 = p, x 4 = e, where p is any parameter.

The described method gives a solution to any system whose augmented matrix is in the RREF. Solving systems whose augmented matrix is in the REF is very similar, but there is a little bit of extra work.

Gaussian elimination: Elementary operations on matrices Three types of elementary operations on matrices

Gaussian elimination: Elementary operations on matrices Three types of elementary operations on matrices 1. Type I: Interchange two rows of a matrix. 2. Type II: Multiply a row by a nonzero number. 3. Type III: Add a multiple of one row to a different row. Theorem Every matrix can be brought to REF (or RREF) by a sequence of elementary row operations.

A proof of the theorem

A proof of the theorem Let s see how this works in practice.

Putting it all together